2023 AMC 8 Problems/Problem 6

Revision as of 22:01, 24 January 2023 by Apex304 (talk | contribs)

Solution 1

First, let us consider the cases where $0$ is a base. This would result in the entire expression being $0$. However, if $0$ is an exponent, we will get a value greater than $0$. As $3^2\cdot2^0=9$ is greater than $2^3\cdot2^0=8$ and $2^2\cdot3^0=4$, the answer is $\boxed{\textbf{(C) }9}$.

~MathFun1000

Solution 2

The maximum possible value of using the digit $2,0,2,3$. We can maximize our value by keeping the $3$ and $2$ together in one power. (Biggest with biggest and smallest with smallest) This shows $3^{2}*0^{2}$=$9*1$=$9$. (Don't want $2^{0}$ cause that's $0$) It is going to be $\boxed{\text{(C)}9}$

~apex304 (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209 (editing))